# How is a Fock state from QFT related to the wave function from quantum mechanics?

I am currently studying quantum field theory as part of my degree. I'm just lacking intuition or an understanding of some basic concepts. So please don't hesitate to correct me if i got something wrong.

Let's start with the quantization of the real scalar field. As I understand it, we divide the field operator into ladder operators. We can set up commutation relations for these, which we get from the commutation relations of the field operator and the conjugate momentum $$[\psi(\vec x), \pi(\vec y)] = i \delta^3(\vec x - \vec y)$$ (which we assume?). The commutation relations of the ladder operators then lead to the existence of discrete energy eigenstates, which we call particles (similar to the harmonic oscillator in QM). With the ladder operators, we can now generate particles from the ground state/vacuum and evolve them with the Hamiltonian $$H$$ the same way as in QM.

That's my understanding so far. Similar to quantum mechanics, where the wave function of a free particle can be represented as the superposition of different momentum eigenstates, I initially thought that this is also the case for Fock states. In this case, the momentum eigenstates would be defined by the creation operator of the respective momentum:

QM QFT
$$|\psi\rangle = \int \frac{\mathrm{d}^3p}{(2\pi)^3} \tilde \psi(\vec p) |\vec p\rangle$$ $$|\psi\rangle = \int \frac{\mathrm{d}^3p}{(2\pi)^3} \tilde \psi(\vec p) a^\dagger(\vec p)|0\rangle$$

The norms of the two states are the same. With real scalar fields this works quite well, but in spinor fields my understanding of the meaning of the state stops. In quantum mechanics, our wave function is no longer described with a complex number, but with a spinor. Logically, so is the quantum field operator $$\psi(x)$$. But not the state. There, the state is still described by combinations of the creation and annihilation operators, just as in the real case.

It is clear to me that this must be the case; after all, it is still a state vector. My problem is just the connection with the wave function, which is no longer tangible for me from here on. It was still simple with the scalar field, where we had a creation operator weighted by the wave function.

My question now is: How do I get from a Fock state (for example $$a_s^\dagger(\vec p)|0\rangle$$) to the corresponding spinor wave function? I am missing this connection.

The 4 spinor solutions of the Dirac equation $$u_1(\vec p)$$, $$u_2(\vec p)$$, $$v_1(\vec p)$$ and $$v_2(\vec p)$$ do not occur anywhere in the state, but only in the field operator, which is not used for describing fock states. In some other lectures some calculations were done with this solutions and i don't get where these show up in QFT. I get how QFT describes the evolution, creation and annihilation of particles but i have no intuitive understanding of the fock state in contrast to QM, where $$\psi^\dagger(\vec x)\psi(\vec x)$$ is clearly defined as the probability density for the particle to be at $$\vec x$$.

One thing that also confuses me is the complex phase after time evolution. One thing that is striking for antiparticles is that the spinor wave function does not evolve as usual with $$e^{-i \omega t}$$ but with $$e^{i \omega t}$$. This follows directly from the Dirac equation. However, this is not the case for the Fock state. Both particles and antiparticles have the same eigenvalues for the free Hamiltonian. The consequence of this is an evolution of $$e^{-i \omega t}$$ for both particles. This does not even require the use of a spinor field. The antiparticles of a complex scalar field behave in the same way.

In QM, the wave function $$\psi(x)$$ provides a complete description of the state of a particle. In QFT, the field operator $$\hat{\psi}(x)$$, which is a solution to the Dirac equation, plays a similar role. However, it operates on the vacuum (or other states) to create or annihilate particle states. The field operator is expanded in terms of creation and annihilation operators, each weighted by the solutions of the Dirac equation (the spinor solutions $$u_s(p)$$ and $$v_s(p)$$). The Fock state, such as $$a^\dagger_s(p)|0\rangle$$, represents a single particle state with momentum $$p$$ and spin $$s$$. To connect this with the spinor wave function, consider the field operator's action. For instance, $$\hat{\psi}(x)$$ acting on the vacuum can create a particle at position $$x$$, and its expression involves the spinor solutions $$u_s(p)$$ and $$v_s(p)$$ integrated over all momenta, reflecting the superposition principle.
In QM, the probability density is given by $$\psi^\dagger(x)\psi(x)$$, which has a clear physical interpretation. In QFT, the analogous quantity is the expectation value of the number operator $$N = a^\dagger a$$, which gives the number of particles in a given state. The field operator $$\hat{\psi}(x)$$ in QFT, when acting on the vacuum, creates a particle at position $$x$$, but the interpretation is subtly different due to the theory's many-body nature. The time evolution of states in QFT follows from the Hamiltonian. For antiparticles, the reversal in the sign of the time-evolution phase arises from the properties of the antiparticle solutions in the Dirac equation. However, in the Fock space description, both particles and antiparticles evolve with the usual $$e^{-i\omega t}$$ phase. This apparent discrepancy is resolved by understanding that the creation and annihilation operators for particles and antiparticles are distinct, and the time evolution of states in Fock space includes both particle and antiparticle contributions, each evolving according to their respective Hamiltonian terms.